選擇在mysql中重復的字段

Question

假設我有下面的customer_offer表。

我的問題是：

如何選擇該表中所有重復鍵的行？

+---------+-------------+------------+----------+--------+---------------------+
| link_id | customer_id | partner_id | offer_id | key    | date_updated        |
+---------+-------------+------------+----------+--------+---------------------+
|       1 |          99 |         11 |       14 | mmmmmq | 2011-09-21 12:40:46 |
|       2 |         100 |         11 |       14 | qmmmmq | 2011-09-21 12:40:46 |
|       3 |         101 |         11 |       14 | 8mmmmq | 2011-09-21 12:40:46 |
|       4 |          99 |         11 |       14 | Dmmmmq | 2011-09-21 12:59:28 |
|       5 |         100 |         11 |       14 | Nmmmmq | 2011-09-21 12:59:28 |
+---------+-------------+------------+----------+--------+---------------------+

更新：

非常感謝您的所有答復。 有很多答案是好的。 現在我有解決方案了。

Answer 1

select * 
from customer_offer 
where key in 
(select key from customer_offer group by key having count(*) > 1)

更新：如@ Scorpi0所述，如果表很大，最好使用join。 從mysql6.0開始，新的優化器會將這種子查詢轉換為聯接。

Answer 2

Self join

SELECT * FROM customer_offer c1 inner join customer_offer c2
on c1.key = c2.key

或group by字段group by然后在count > 1時取

SELECT COUNT(key),link_id FROM customer_offer c1
group by key, link_id
having COUNT(Key) > 1

Answer 3

SELECT DISTINCT c1.* 
FROM customer_offer c1 
    INNER JOIN customer_offer c2
        ON c1.key = c2.key
        AND c1.link_id != c2.link_id

假設link_id是主鍵。

Answer 4

mysql網站上有許多線程介紹了如何執行此操作。 該鏈接將說明如何使用mysql執行此操作： http : //forums.mysql.com/read.php?10,180556,180567#msg-180567

作為一個簡短的示例，下面的代碼來自鏈接，並進行了一些細微修改，從而更適合您的示例。

SELECT *
FROM tbl
GROUP BY key
HAVING COUNT(key)>1; 

您也可以使用joing，這是我的首選方法，因為這消除了計數較慢的方法：

SELECT * 
FROM this_table t 
inner join this_table t1 on t.key = t1.key

Answer 5

使用子查詢進行計數檢查，並使用主查詢選擇行。 計數檢查查詢很簡單：

SELECT `link_id` FROM `customer_offer` GROUP BY `key` HAVING COUNT(`key`) > 1

然后外部查詢將通過加入它來使用它：

SELECT customer_offer.* FROM customer_offer
INNER JOIN (SELECT `link_id` FROM `customer_offer` GROUP BY `key` HAVING COUNT(`key`) > 1) AS count_check
ON customer_offer.link_id = count_check.link_id

Answer 6

SELECT link_id, key, count(key) as Occurrences 
FROM table 
GROUP BY key
HAVING COUNT(key)>1;